The given situation can be shown as :

Distance between H and Cl atoms = 1.27 \(\AA\)
Mass of H atom = \(m\)
Mass of Cl atom = \(35.5\, m\)
Let the centre of mass of the system lie at a distance x from the CI atom
Distance of the centre of mass from the H atom = (\(1.27-x\))
Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
\(\frac{m(1.27-x)+35.5mx}{m+35.5m}\) = \(0\)
\(m(1.27-x)+35.5\,mx\) = \(0\)
\(1.27-x\) = \(35.5x\)
\(\therefore\) \(x\) = \(\frac{-1.27}{(35.5-1)}\) = \(-0.037 \AA\)
Here, the negative sign indicates that the centre of mass lies at the left of the molecule.
Hence, the centre of mass of the HCl molecule lies \(0.037 \AA\) from the Cl atom.


Find the value of m if \(M = 10\) \(kg\). All the surfaces are rough.
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
